Simplify and expand the following expression: $ \dfrac{3}{5t + 40}+ \dfrac{1}{2t - 8}- \dfrac{1}{t^2 + 4t - 32} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{3}{5t + 40} = \dfrac{3}{5(t + 8)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{1}{2t - 8} = \dfrac{1}{2(t - 4)}$ We can factor the quadratic in the third term: $ \dfrac{1}{t^2 + 4t - 32} = \dfrac{1}{(t + 8)(t - 4)}$ Now we have: $ \dfrac{3}{5(t + 8)}+ \dfrac{1}{2(t - 4)}- \dfrac{1}{(t + 8)(t - 4)} $ The least common multiple of the denominators is: $ 10(t + 8)(t - 4)$ In order to get the first term over $10(t + 8)(t - 4)$ , multiply by $\dfrac{2(t - 4)}{2(t - 4)}$ $ \dfrac{3}{5(t + 8)} \times \dfrac{2(t - 4)}{2(t - 4)} = \dfrac{6(t - 4)}{10(t + 8)(t - 4)} $ In order to get the second term over $10(t + 8)(t - 4)$ , multiply by $\dfrac{5(t + 8)}{5(t + 8)}$ $ \dfrac{1}{2(t - 4)} \times \dfrac{5(t + 8)}{5(t + 8)} = \dfrac{5(t + 8)}{10(t + 8)(t - 4)} $ In order to get the third term over $10(t + 8)(t - 4)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{1}{(t + 8)(t - 4)} \times \dfrac{10}{10} = \dfrac{10}{10(t + 8)(t - 4)} $ Now we have: $ \dfrac{6(t - 4)}{10(t + 8)(t - 4)} + \dfrac{5(t + 8)}{10(t + 8)(t - 4)} - \dfrac{10}{10(t + 8)(t - 4)} $ $ = \dfrac{ 6(t - 4) + 5(t + 8) - 10} {10(t + 8)(t - 4)} $ Expand: $ = \dfrac{6t - 24 + 5t + 40 - 10}{10t^2 + 40t - 320} $ $ = \dfrac{11t + 6}{10t^2 + 40t - 320}$